---
title: "NCERT Solutions Class 8 Maths The Baudhayana-Pythagoras Theorem"
url: https://www.swavid.com/maths/class/8/chapter/the-baudhayana-pythagoras-theorem/ncert-solutions
dateModified: 2026-10-07T15:35:11+00:00
---

# NCERT Solutions Class 8 Maths The Baudhayana-Pythagoras Theorem

This chapter's questions cover the geometric concepts of doubling and halving squares, understanding the hypotenuse of isosceles right triangles, and applying the Baudhayana-Pythagoras theorem to various problems.

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## 2.1 Doubling a Square

### Question 1

*3 marks · Short answer*

How can one construct a square having double the area of a given square?

**Solution**

1. Construct a square on the diagonal of the given square.
2. The new square is made up of four small triangles, whereas the original square is made up of two such congruent small triangles.
3. Therefore, the diagonal of a square produces a square of double the area of the original square.

**Answer:** Construct a square on the diagonal of the given square to get a square with double the area.

> Common mistake: Thinking that doubling the length of each side doubles the area, instead of quadrupling it.

## Page 34 In-text Questions

### Question 1

*3 marks · Short answer*

Why does the new dotted square have double the area of the original square?

**Solution**

1. The original square can be divided into two congruent right-angled isosceles triangles by its diagonal.
2. The new dotted square is constructed on the diagonal of the original square.
3. Each small triangle has an area equal to half of the original square, and the new square is made up of four such small triangles, giving it double the area.

**Answer:** The new dotted square has double the area because it is made up of four small triangles while the original square contains two.

> Common mistake: Assuming that doubling the sidelength doubles the area instead of quadrupling it.

### Question 2

*3 marks · Short answer*

Can you draw some horizontal and vertical lines to see why the new square has double the area of the original square?

**Solution**

1. Extend the horizontal and vertical sides (east-west and north-south lines) of the original square across the dotted square.
2. These perpendicular lines divide the squares into smaller regions or triangles.
3. Counting the triangles shows that the original square consists of two small triangles, whereas the dotted square consists of four identical small triangles.

**Answer:** Drawing horizontal and vertical lines divides the original square into 2 small triangles and the dotted square into 4 small triangles, showing the area is doubled.

> Common mistake: Drawing lines incorrectly without aligning them with the sides and diagonals of the square.

### Question 3

*3 marks · Short answer*

Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square?

**Solution**

1. The diagonal of a square bisects its interior angles.
2. By symmetry and the diagonal property of a square, the extension of the vertical and horizontal sides of the original square pass through the opposite vertices of the dotted square.
3. These extension lines act as angle bisectors of the corners of the dotted square.

**Answer:** The extension lines pass through the vertices because they bisect the angles of the dotted square due to the diagonal property.

> Common mistake: Forgetting that the sides of the original square lie along the lines of symmetry of the constructed figure.

### Question 4

*3 marks · Short answer*

Moreover, all these small triangles are congruent to each other. Can you explain why?

**Solution**

1. The horizontal and vertical lines divide the squares into right-angled triangles.
2. Each small triangle has sides formed by the radius or half-side segments of the grid with a $90^\circ$ angle between them.
3. By matching their sides and interior angles, all these small triangles are congruent to each other.

**Answer:** All the small triangles are congruent because they have corresponding equal sides and right angles formed by the grid lines.

> Common mistake: Failing to verify the side lengths and right angles of the constituent triangles.

## Page 35 In-text Questions

### Question 1

*Activity*

Cut out two identical squares of paper. Draw, label, and cut as follows: Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square with double the area.

**Solution**

1. Cut out two identical squares of paper and label their triangular sections as shown in the textbook (Fig. on page 35).
2. Place pieces 5, 6, 7, and 8 around Square 1 (which consists of triangles 1, 2, 3, and 4) to form a larger square.
3. Observation: The newly formed square is composed of 8 small congruent triangles, whereas the original square contains 4 small triangles, thus achieving a square with double the area.

**Answer:** The pieces 5, 6, 7, and 8 placed around Square 1 form a square of double the area.

## 2.2 Halving a Square

### Question 1

*3 marks · Short answer*

Now suppose we are given a square, and we want to construct a square whose area is half that of the original square. How would you do it?

**Solution**

1. We are given a square and want to construct a square whose area is half of it.
2. We can do this by drawing a tilted smaller square inside the larger square.
3. The vertices of the smaller square lie on the midpoints of the sides of the original square.

**Answer:** We construct a tilted smaller square inside the larger square by joining the midpoints of its sides.

> Common mistake: Thinking that halving the sidelength halves the area instead of halving the area by connecting midpoints.

### Question 2

*3 marks · Short answer*

Why is the smaller inside square half the area of the larger square? Again, adding some east-west and north-south lines can explain it:

**Solution**

1. Adding horizontal (east-west) and vertical (north-south) lines through the vertices of the inner square divides the larger square into eight congruent right-angled triangles.
2. The original square is made up of eight such small triangles, while the inner square is made up of four such small triangles.
3. Since the inner square contains four small triangles and the larger square contains eight, the area of the inner square is half the area of the larger square.

**Answer:** The smaller square is made up of four small triangles while the larger square is made up of eight, so its area is half.

> Common mistake: Counting the triangles incorrectly.

### Question 3

*Activity*

Cut out a square from a piece of paper. Now make a square whose area is half the area of the first square.

**Solution**

1. Fold the square paper inward such that the crease lines pass through the midpoints of the sides.
2. The resulting quadrilateral PQRS is a square whose area is half the area of the original square paper.

**Answer:** Folding the paper along the midpoints of its sides gives a square with half the area.

### Question 4

*3 marks · Short answer*

Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?

**Solution**

1. No, a square having half the sidelength will not have half the area.
2. If the sidelength is halved, the area becomes $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the original area.
3. Four such squares with half the sidelength will fill the original square.

**Answer:** No, halving the sidelength makes the area one-fourth. Four such squares will fill the original square.

> Common mistake: Assuming area scales linearly with sidelength.

## Page 37 In-text Questions

### Question 1

*3 marks · Short answer*

Why is PQRS a square? Why is its area half that of the original paper? Explain by connecting QS and PR, finding the different angles formed, and then using tringle congruence.

**Solution**

1. 1. When the square paper is folded inward such that crease lines pass through the midpoints of the sides, it forms a smaller inner quadrilateral PQRS.
2. 2. Connecting the diagonals QS and PR divides the squares into triangles where corresponding triangles are congruent by SAS criterion, showing all sides of PQRS are equal and angles are $90^\circ$.
3. 3. Since the inner square is made of half the triangles of the original square, its area is exactly half the area of the original paper.

**Answer:** PQRS is a square because its sides are equal and angles are $90^\circ$, and its area is half of the original paper because it contains half the component triangles.

> Common mistake: Confusing the area scaling factor with side length scaling factor.

### Question 2

*3 marks · Short answer*

Find the hypotenuse of this isosceles right triangle.

**Solution**

1. 1. The area of the square constructed on the hypotenuse is twice the area of the square made by the two equal sides, so Area = $2 \times 1 = 2 \text{ sq. units}$.
2. 2. If $c$ is the length of the hypotenuse, then the area of the square on it is $c^2 = 2$.
3. 3. Taking the square root gives $c = \sqrt{2}$, so the length of the hypotenuse is $\sqrt{2}$ units.

**Answer:** $\sqrt{2}$ units

> Common mistake: Writing the hypotenuse length as 2 instead of $\sqrt{2}$.

## Page 38 In-text Questions

### Question 1

*3 marks · Short answer*

What is the value of $\sqrt{2}$?

**Solution**

1. The exact value of $\sqrt{2}$ cannot be expressed as a terminating decimal or a simple fraction.
2. Using successive approximations and bounds, the decimal representation is non-terminating.
3. Thus, the value of $\sqrt{2}$ is approximately $1.41421356\dots$.

**Answer:** $1.41421356\dots$

> Common mistake: Writing a terminating decimal like $1.414$ as the exact value of $\sqrt{2}$.

### Question 2

*3 marks · Short answer*

Is $\sqrt{2}$ less than or greater than 1?

**Solution**

1. A square of sidelength $1$ unit has an area of $1$ sq. unit.
2. A square of sidelength $\sqrt{2}$ has an area of $2$ sq. units, since $1^2 = 1$ and $(\sqrt{2})^2 = 2$.
3. Since $1 < 2$, $\sqrt{2}$ is greater than $1$, meaning $1 < \sqrt{2}$.

**Answer:** $\sqrt{2}$ is greater than 1 ($1 < \sqrt{2}$)

> Common mistake: Confusing the areas of the squares with their sidelengths.

### Question 3

*3 marks · Short answer*

Is $\sqrt{2}$ less than or greater than 2?

**Solution**

1. A square of sidelength $2$ units has an area of $4$ sq. units.
2. A square of sidelength $\sqrt{2}$ has an area of $2$ sq. units, since $(\sqrt{2})^2 = 2$ and $2^2 = 4$.
3. Since $2 < 4$, $\sqrt{2}$ is less than $2$, meaning $\sqrt{2} < 2$.

**Answer:** $\sqrt{2}$ is less than 2 ($\sqrt{2} < 2$)

> Common mistake: Comparing the numbers incorrectly by forgetting to square them.

### Question 4

*3 marks · Short answer*

Can we find closer bounds for $\sqrt{2}$?

**Solution**

1. We test tenths and find that $1.4^2 = 1.96$ and $1.5^2 = 2.25$, so $1.4 < \sqrt{2} < 1.5$.
2. We test hundredths and find that $1.41^2 = 1.9881$ and $1.42^2 = 2.0164$, so $1.41 < \sqrt{2} < 1.42$.
3. We can continue this process to find arbitrarily close bounds like $1.414 < \sqrt{2} < 1.415$.

**Answer:** Yes, by testing decimal squares we get tighter bounds such as $1.414 < \sqrt{2} < 1.415$

> Common mistake: Arithmetic errors when squaring decimals like $1.41$ and $1.42$.

### Question 5

*3 marks · Short answer*

Will we ever get a number with a terminating decimal representation whose square is 2?

**Solution**

1. If there were a terminating decimal whose square is $2$, its last decimal digit after squaring would also be non-zero.
2. For example, if the decimal ended in $4$, its square would end in $6$, which cannot equal $2$ (or $2.000\dots$).
3. Thus, a terminating decimal cannot have $2$ as its square, meaning $\sqrt{2}$ has a non-terminating decimal representation.

**Answer:** No, a terminating decimal can never have 2 as its square.

> Common mistake: Assuming that decimals must terminate because we can only write a finite number of digits on paper.

## Page 39 In-text Questions

### Question 1

*3 marks · Short answer*

Can $\sqrt{2}$ be expressed as a fraction $\frac{m}{n}$, where $m$ and $n$ are counting numbers?

**Solution**

1. Assume to the contrary that $\sqrt{2}$ can be expressed as a fraction $\frac{m}{n}$ where $m$ and $n$ are counting numbers.
2. Squaring both sides gives $2 = \frac{m^2}{n^2}$, which implies $2n^2 = m^2$.
3. In the prime factorization of a square number, each prime occurs an even number of times, so $2n^2$ has an odd number of factors of $2$ while $m^2$ has an even number, which is impossible. Thus, $\sqrt{2}$ cannot be expressed as a fraction.

**Answer:** No, $\sqrt{2}$ cannot be expressed as a fraction $\frac{m}{n}$.

> Common mistake: Assuming fractions can always represent square roots of non-square numbers without checking prime factorization properties.

## Figure it Out

### Question 1

*3 marks · Short answer*

Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way. Can you arrange these pieces to create a square with double the area of either square?

**Solution**

1. Take two identical square papers and cut them as shown in the figure into four pieces labeled 1, 2, 3, and 4.
2. Observe that pieces 1, 2, 3, and 4 together make up the area of the two original squares combined.
3. Rearrange these four pieces symmetrically around a central point to form a larger square whose area is double the area of either original square.

**Answer:** The four pieces can be arranged to form a larger square of double the area.

> Common mistake: Trying to fit the pieces without matching the cut edges correctly.

### Question 2

*3 marks · Short answer*

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.
(i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9

**Part (i)**

1. Using the relation for an isosceles right triangle, the hypotenuse is $c = a\sqrt{2}$.
2. Here $a = 3$, so $c = 3\sqrt{2} = \sqrt{18}$ units.
3. Since $4.2^2 = 17.64$ and $4.3^2 = 18.49$, the bounds are $4.2 < \sqrt{18} < 4.3$.

Answer (i): Hypotenuse = $3\sqrt{2}$ units (between $4.2$ and $4.3$)

**Part (ii)**

1. Using the relation for an isosceles right triangle, the hypotenuse is $c = a\sqrt{2}$.
2. Here $a = 4$, so $c = 4\sqrt{2} = \sqrt{32}$ units.
3. Since $5.6^2 = 31.36$ and $5.7^2 = 32.49$, the bounds are $5.6 < \sqrt{32} < 5.7$.

Answer (ii): Hypotenuse = $4\sqrt{2}$ units (between $5.6$ and $5.7$)

**Part (iii)**

1. Using the relation for an isosceles right triangle, the hypotenuse is $c = a\sqrt{2}$.
2. Here $a = 6$, so $c = 6\sqrt{2} = \sqrt{72}$ units.
3. Since $8.4^2 = 70.56$ and $8.5^2 = 72.25$, the bounds are $8.4 < \sqrt{72} < 8.5$.

Answer (iii): Hypotenuse = $6\sqrt{2}$ units (between $8.4$ and $8.5$)

**Part (iv)**

1. Using the relation for an isosceles right triangle, the hypotenuse is $c = a\sqrt{2}$.
2. Here $a = 8$, so $c = 8\sqrt{2} = \sqrt{128}$ units.
3. Since $11.3^2 = 127.69$ and $11.4^2 = 129.96$, the bounds are $11.3 < \sqrt{128} < 11.4$.

Answer (iv): Hypotenuse = $8\sqrt{2}$ units (between $11.3$ and $11.4$)

**Part (v)**

1. Using the relation for an isosceles right triangle, the hypotenuse is $c = a\sqrt{2}$.
2. Here $a = 9$, so $c = 9\sqrt{2} = \sqrt{162}$ units.
3. Since $12.7^2 = 161.29$ and $12.8^2 = 163.84$, the bounds are $12.7 < \sqrt{162} < 12.8$.

Answer (v): Hypotenuse = $9\sqrt{2}$ units (between $12.7$ and $12.8$)

**Answer:** Calculated the hypotenuse and its bounds with one decimal digit for each given side length.

> Common mistake: Forgetting to square the side length before multiplying by 2 when converting to a single square root.

## Page 40 In-text Questions

### Question 3

*3 marks · Short answer*

The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]

**Solution**

1. Let $a$ be the length of the equal sides and $c = 10$ be the length of the hypotenuse.
2. Using the general relation for an isosceles right triangle, $c^2 = 2a^2$.
3. Substitute $c = 10$ into the formula to get $10^2 = 2a^2$, which gives $100 = 2a^2$ or $a^2 = 50$.
4. Therefore, each of the other two sides has length $a = \sqrt{50} = 5\sqrt{2}$ units.

**Answer:** $5\sqrt{2}$ units

> Common mistake: Dividing the hypotenuse by 2 instead of using the relation $c^2 = 2a^2$.

## Page 41 In-text Questions

### Question 1

*3 marks · Short answer*

What if we wish to combine two squares of ‘different’ sizes to make a large square whose area is the sum of the two smaller squares?

**Solution**

1. Make a right-angled triangle whose perpendicular sides are equal to the sidelengths of the two given squares.
2. Construct a square on the hypotenuse of this right-angled triangle.
3. By Baudhāyana's theorem, the area of this new square equals the sum of the areas of the two original squares.

**Answer:** Make a right-angled triangle with perpendicular sides equal to the sidelengths of the two squares; the square on its hypotenuse has an area equal to the sum of the areas of the two squares.

> Common mistake: Trying to simply add the sidelengths instead of constructing a right triangle and taking the square on the hypotenuse.

## Page 42 In-text Questions

### Question 1

*3 marks · Short answer*

Why does Baudhāyana’s method work?

**Solution**

1. When we make a right-angled triangle whose perpendicular sides are the sidelengths $a$ and $b$ of two given squares, the square on the hypotenuse $c$ satisfies $c^2 = a^2 + b^2$ by Baudhayana's theorem.
2. By arranging four such right-angled triangles around a central square of side $(b-a)$, a larger square of side $c$ is formed.
3. The area of this new square equals the sum of the areas of the four right triangles and the inner square, which simplifies to $a^2 + b^2$, thus combining the areas of the two given squares.

**Answer:** Baudhayana's method works because arranging four congruent right-angled triangles of sides $a$ and $b$ around an inner square forms a larger square of side $c$ whose area equals the sum of the areas of the two original squares ($a^2 + b^2 = c^2$).

> Common mistake: Confusing the areas of the individual triangles with the area of the combined square.

### Question 2

*3 marks · Short answer*

Can you see why the method works in the case where the two squares are the same size? Does it agree with the method we used earlier to combine two same sized squares into a bigger square?

**Solution**

1. When the two squares are of the same size, their sidelengths are both equal to $a$, so the perpendicular sides of the right-angled triangle are $a$ and $a$.
2. The square on the hypotenuse $c$ satisfies $c^2 = a^2 + a^2 = 2a^2$.
3. This is identical to the method used earlier where the diagonal of a square of side $a$ produces a square of double the area ($2a^2$), completely agreeing with both approaches.

**Answer:** Yes, when both squares are the same size, the perpendicular sides are equal, and the hypotenuse becomes the diagonal of the square, which matches the earlier method of doubling a square.

> Common mistake: Failing to substitute $b = a$ into the general formula.

## Page 44 In-text Questions

### Question 1

*3 marks · Short answer*

Why?

**Solution**

1. The 4-sided figure is formed by four right-angled triangles T, U, X, and W placed around a central square V.
2. Each of these triangles has perpendicular sides of lengths $a$ and $b$, and hypotenuse equal to the side of the new 4-sided figure.
3. Since all four triangles are congruent by SAS congruency, their hypotenuses are equal, making all four sides of the 4-sided figure equal in length.

**Answer:** The sides of the new 4-sided figure are all equal because they are formed by the hypotenuses of four congruent right-angled triangles.

> Common mistake: Stating that the sides are equal without mentioning the congruency of the triangles.

### Question 2

*3 marks · Short answer*

Explain why all the angles of this new 4-sided figure are right angles and so it is a square.

**Solution**

1. Let the acute angles of the right-angled triangle with sides $a$ and $b$ be $x$ and $90^\circ - x$.
2. At each vertex of the new 4-sided figure, the angle is made up of the sum of the two acute angles $x$ and $90^\circ - x$ of the adjacent right triangles along a straight line.
3. Since $x + (90^\circ - x) = 90^\circ$, each interior angle of the 4-sided figure is a right angle, and therefore the figure is a square.

**Answer:** Each angle of the new 4-sided figure is $90^\circ$ because it is composed of the two complementary acute angles of the right-angled triangles, making it a square.

> Common mistake: Forgetting to show that the sum of the non-right angles at the vertices equals $90^\circ$.

## Figure it Out

### Question 1

*3 marks · Short answer*

If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana’s Theorem.

**Solution**

1. Let the shorter sides be $a = 5\text{ cm}$ and $b = 12\text{ cm}$, and let the hypotenuse be $c$.
2. By Baudhāyana's Theorem, $a^2 + b^2 = c^2$, so $5^2 + 12^2 = c^2$.
3. This gives $25 + 144 = c^2$, which means $c^2 = 169$, so $c = 13\text{ cm}$.
4. The length of the hypotenuse is $13\text{ cm}$.

**Answer:** 13 cm

> Common mistake: Adding the lengths of the sides directly instead of squaring them.

### Question 2

*3 marks · Short answer*

If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana’s Theorem.

**Solution**

1. Let the short side be $a = 8\text{ cm}$, the hypotenuse be $c = 17\text{ cm}$, and the third side be $b$.
2. By Baudhāyana's Theorem, $a^2 + b^2 = c^2$, so $8^2 + b^2 = 17^2$.
3. This gives $64 + b^2 = 289$, so $b^2 = 289 - 64 = 225$.
4. Therefore, $b = 15\text{ cm}$.

**Answer:** 15 cm

> Common mistake: Adding the squares of the short side and hypotenuse instead of subtracting.

### Question 3

*3 marks · Short answer*

Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhāyana’s Śulba-Sūtra, Verse 1.10)

**Solution**

1. To construct a square of triple the area, construct a right triangle whose sides are the side of the original square and the diagonal of the square, and draw a square on its hypotenuse.
2. The hypotenuse squared will be $1^2 + (\sqrt{2})^2 = 3$ times the area of the original square.
3. To get five times the area, use a right triangle with sides equal to the side of the original square and the diagonal of a square with double the area ($2$ units), or use sides of lengths $2$ and $1$ units whose hypotenuse squared is $2^2 + 1^2 = 5$.

**Answer:** Construct a square on the hypotenuse of a right-angled triangle with appropriate side lengths using Baudhāyana's method.

> Common mistake: Multiplying the side length by 3 or 5 instead of using the diagonal construction.

### Question 4

*3 marks · Short answer*

Let $a$, $b$ and $c$ denote the length of the sides of a right triangle, with $c$ being the length of the hypotenuse. Find the missing sidelength in each of the following cases:
(i) $a = 5, b = 7$
(ii) $a = 8, b = 12$
(iii) $a = 9, c = 15$
(iv) $a = 7, b = 12$
(v) $a = 1.5, b = 3.5$

**Part (i)**

1. We have $a = 5$ and $b = 7$, and we need to find $c$.
2. Using $c^2 = a^2 + b^2$, we get $c^2 = 5^2 + 7^2 = 25 + 49 = 74$.
3. Thus, $c = \sqrt{74}$.

Answer (i): \sqrt{74}

**Part (ii)**

1. We have $a = 8$ and $b = 12$, and we need to find $c$.
2. Using $c^2 = a^2 + b^2$, we get $c^2 = 8^2 + 12^2 = 64 + 144 = 208$.
3. Thus, $c = \sqrt{208}$.

Answer (ii): \sqrt{208}

**Part (iii)**

1. We have $a = 9$ and $c = 15$, and we need to find $b$.
2. Using $a^2 + b^2 = c^2$, we get $9^2 + b^2 = 15^2$, so $81 + b^2 = 225$.
3. Thus, $b^2 = 225 - 81 = 144$, which gives $b = 12$.

Answer (iii): 12

**Part (iv)**

1. We have $a = 7$ and $b = 12$, and we need to find $c$.
2. Using $c^2 = a^2 + b^2$, we get $c^2 = 7^2 + 12^2 = 49 + 144 = 193$.
3. Thus, $c = \sqrt{193}$.

Answer (iv): \sqrt{193}

**Part (v)**

1. We have $a = 1.5$ and $b = 3.5$, and we need to find $c$.
2. Using $c^2 = a^2 + b^2$, we get $c^2 = (1.5)^2 + (3.5)^2 = 2.25 + 12.25 = 14.5$.
3. Thus, $c = \sqrt{14.5}$.

Answer (v): \sqrt{14.5}

**Answer:** Calculated the missing sidelength for each case using Baudhāyana's Theorem.

> Common mistake: Mixing up the hypotenuse and shorter sides when applying the theorem.

## Page 48 In-text Questions

### Question 1

*3 marks · Short answer*

List down all the Baudhāyana triples with numbers less than or equal to 20.

**Solution**

1. A Baudhāyana triple is a triple of positive integers $(a, b, c)$ that satisfy the relation $a^2 + b^2 = c^2$.
2. The Baudhāyana triples with numbers less than or equal to 20 listed in the chapter are $(3, 4, 5)$, $(5, 12, 13)$, $(6, 8, 10)$, $(8, 15, 17)$, $(9, 12, 15)$, and $(12, 16, 20)$.
3. Thus, the required Baudhāyana triples with numbers $\le 20$ are $(3, 4, 5)$, $(5, 12, 13)$, $(6, 8, 10)$, $(8, 15, 17)$, $(9, 12, 15)$, and $(12, 16, 20)$.

**Answer:** $$(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20)$$

> Common mistake: Forgetting scaled versions like (6, 8, 10) or (9, 12, 15) which also have numbers $\le 20$.

### Question 2

*3 marks · Short answer*

Is there an unending sequence of Baudhāyana triples?

**Solution**

1. Yes, there is an unending sequence of Baudhāyana triples.
2. Mathematicians have found methods to generate infinitely many such triples, such as by taking scaled versions $(ka, kb, kc)$ for any positive integer $k$.
3. Thus, the sequence of Baudhāyana triples is unending.

**Answer:** Yes, there is an unending sequence of Baudhāyana triples.

> Common mistake: Thinking that only a fixed list of triples exists.

### Question 3

*3 marks · Short answer*

Is (30, 40, 50) a Baudhāyana triple? Is (300, 400, 500) a Baudhāyana triple?

**Solution**

1. For $(30, 40, 50)$, we check if $30^2 + 40^2 = 50^2$.
2. We have $900 + 1600 = 2500$, which equals $50^2$. So $(30, 40, 50)$ is a Baudhāyana triple.
3. Similarly, for $(300, 400, 500)$, $300^2 + 400^2 = 90000 + 160000 = 250000 = 500^2$. So $(300, 400, 500)$ is also a Baudhāyana triple.

**Answer:** Yes, both $(30, 40, 50)$ and $(300, 400, 500)$ are Baudhāyana triples.

> Common mistake: Calculation errors while squaring larger numbers.

### Question 4

*3 marks · Short answer*

Do you see any pattern among them?

**Solution**

1. Consider the triples $(3, 4, 5)$, $(6, 8, 10)$, $(9, 12, 15)$, and $(12, 16, 20)$.
2. Notice that $(6, 8, 10) = 2 \times (3, 4, 5)$, $(9, 12, 15) = 3 \times (3, 4, 5)$, and $(12, 16, 20) = 4 \times (3, 4, 5)$.
3. All these triples are obtained by multiplying each term of the base triple $(3, 4, 5)$ by a positive integer.

**Answer:** All these triples are obtained by multiplying each term of $(3, 4, 5)$ by a positive integer.

> Common mistake: Failing to recognize common factors relating the triples.

### Question 5

*3 marks · Short answer*

Can we form a conjecture on Baudhāyana triples based on this observation?

**Solution**

1. Based on the observation that the triples are multiples of $(3, 4, 5)$, we can form a general conjecture.
2. The conjecture states that $(3k, 4k, 5k)$ is a Baudhāyana triple for any positive integer $k$.
3. This gives a systematic way to generate new triples from a known one.

**Answer:** Conjecture: $(3k, 4k, 5k)$ is a Baudhāyana triple, where $k$ is any positive integer.

> Common mistake: Omitting the condition that $k$ must be a positive integer.

### Question 6

*3 marks · Short answer*

Is this true?

**Solution**

1. To check if $(3k, 4k, 5k)$ is a Baudhāyana triple, we need to verify whether $(3k)^2 + (4k)^2 = (5k)^2$.
2. We expand the left side: $(3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$.
3. Since $25k^2 = (5k)^2$, the relation holds true, proving that $(3k, 4k, 5k)$ is indeed a Baudhāyana triple.

**Answer:** Yes, it is true since $(3k)^2 + (4k)^2 = 25k^2 = (5k)^2$.

> Common mistake: Incorrectly squaring the terms as $3k^2$ instead of $(3k)^2 = 9k^2$.

## Page 49 In-text Questions

### Question 1

*3 marks · Short answer*

If $(a, b, c)$ is a Baudhāyana triple, then $(ka, kb, kc)$ is also a Baudhāyana triple where $k$ is any positive integer. Is this statement true?

**Solution**

1. Given that $(a, b, c)$ is a Baudhāyana triple, we have $a^2 + b^2 = c^2$.
2. We need to check whether $(ka)^2 + (kb)^2 = (kc)^2$ for any positive integer $k$.
3. Expanding both sides, we get $(ka)^2 + (kb)^2 = k^2 a^2 + k^2 b^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2$.
4. Therefore, the statement is true.

**Answer:** Yes, the statement is true.

> Common mistake: Forgetting to square the scaling factor $k$ on all terms.

### Question 2

*3 marks · Short answer*

Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive Baudhāyana triples with numbers less than or equal to 20?

**Solution**

1. A Baudhāyana triple that does not have any common factor greater than 1 is called a primitive Baudhāyana triple.
2. For $(5, 12, 13)$, the numbers have no common factor greater than 1, so it is a primitive Baudhāyana triple.
3. The other primitive Baudhāyana triples with numbers less than or equal to 20 are $(3, 4, 5)$ and $(8, 15, 17)$.

**Answer:** Yes, $(5, 12, 13)$ is primitive. The other primitive triples with numbers $\le 20$ are $(3, 4, 5)$ and $(8, 15, 17)$.

> Common mistake: Including scaled triples like $(6, 8, 10)$ or $(9, 12, 15)$ in the list of primitive triples.

### Question 3

*3 marks · Short answer*

Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

**Solution**

1. Consider the primitive triples with numbers less than or equal to 20: $(3, 4, 5)$, $(5, 12, 13)$, and $(8, 15, 17)$.
2. Multiplying each by $k = 2, 3, 4, 5, 6$ gives their scaled versions.
3. For example, 5 scaled versions of $(3, 4, 5)$ are $(6, 8, 10)$, $(9, 12, 15)$, $(12, 16, 20)$, $(15, 20, 25)$, and $(18, 24, 30)$.
4. These scaled versions are not primitive because their elements share a common factor greater than 1.

**Answer:** Scaled versions are obtained by multiplying by positive integers; they are not primitive.

> Common mistake: Stating that scaled versions are primitive.

### Question 4

*3 marks · Short answer*

If $(a, b, c)$ is non-primitive, and the integers have $f$ — greater than 1 — as a common factor, then is $(\frac{a}{f}, \frac{b}{f}, \frac{c}{f})$ a Baudhāyana triple? Check this statement for (9, 12, 15). Justify this statement.

**Solution**

1. If $(a, b, c)$ is a non-primitive Baudhāyana triple with common factor $f$, then $a^2 + b^2 = c^2$.
2. Dividing the entire equation by $f^2$, we get $(\frac{a}{f})^2 + (\frac{b}{f})^2 = (\frac{c}{f})^2$.
3. For $(9, 12, 15)$, the common factor is $f = 3$, and dividing by 3 gives $(3, 4, 5)$, which is a Baudhāyana triple since $3^2 + 4^2 = 5^2$.

**Answer:** Yes, $(\frac{a}{f}, \frac{b}{f}, \frac{c}{f})$ is a Baudhāyana triple. For $(9, 12, 15)$, dividing by 3 gives $(3, 4, 5)$.

> Common mistake: Dividing the components by $f$ instead of dividing their squares by $f^2$ in the algebraic justification.

### Question 5

*3 marks · Short answer*

How do we generate more primitive triples?

**Solution**

1. We can generate more primitive triples by using the relation between the sum of consecutive odd numbers and square numbers.
2. Specifically, if the $n$-th odd number ($2n - 1$) is a square number, say $m^2$, we can express it as a sum of two squares equal to another square.
3. This leads to the equation $(n - 1)^2 + (2n - 1) = n^2$, which helps generate Baudhāyana triples.

**Answer:** Primitive triples are generated using the property where the $n$-th odd number is an odd square.

> Common mistake: Confusing primitive triple generation with simple scaling.

### Question 6

*3 marks · Short answer*

For this, we need to know the nth odd number. What is it?

**Solution**

1. The sequence of odd numbers starts as 1, 3, 5, 7, and so on.
2. The first odd number is $2(1) - 1 = 1$, the second is $2(2) - 1 = 3$, and the third is $2(3) - 1 = 5$.
3. Following this pattern, the $n$-th odd number is given by the algebraic expression $2n - 1$.

**Answer:** The $n$-th odd number is $2n - 1$.

> Common mistake: Writing $2n$ instead of $2n - 1$ for an odd number.

## Page 50 In-text Questions

### Question 1

*3 marks · Short answer*

What is the sum of the first $(n - 1)$ odd numbers?

**Solution**

1. We know that the sum of the first $n$ odd numbers is $n^2$, which means $1 + 3 + 5 + \dots + (2n - 1) = n^2$.
2. To find the sum of the first $(n - 1)$ odd numbers, we stop at the $(n - 1)$-th odd number, which is $2(n - 1) - 1$.
3. Thus, the sum of the first $(n - 1)$ odd numbers is $(n - 1)^2$.

**Answer:** $$(n - 1)^2$$

> Common mistake: Confusing the sum of the first $n$ odd numbers with the sum of the first $(n - 1)$ odd numbers.

### Question 2

*3 marks · Short answer*

Could we have obtained this triple using the equation $(n - 1)^2 + (2n - 1) = n^2$?

**Solution**

1. We are given the equation $(n - 1)^2 + (2n - 1) = n^2$.
2. For the first triple generated using $9$, which is the $5^{\text{th}}$ odd number, we substitute $n = 5$ into the equation.
3. This gives $(5 - 1)^2 + 9 = 5^2$, which simplifies to $4^2 + 3^2 = 5^2$.
4. Yes, we could have directly obtained this triple using the given equation.

**Answer:** Yes, we could have obtained this triple using the given equation by substituting $n = 5$.

> Common mistake: Incorrectly substituting the value of $n$ for the corresponding odd number position.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find 5 more Baudhāyana triples using this idea.

**Solution**

1. Take odd squares such as $49$, $81$, $121$, $169$, and $225$ as the nth odd number $2n - 1$.
2. For $2n - 1 = 49$, $n = 25$, giving the triple $(24, 7, 25)$.
3. Similarly, using $81$, $121$, $169$, and $225$ give $(40, 9, 41)$, $(60, 11, 61)$, $(84, 13, 85)$, and $(112, 15, 113)$.
4. Therefore, 5 more Baudhāyana triples are $(24, 7, 25)$, $(40, 9, 41)$, $(60, 11, 61)$, $(84, 13, 85)$, and $(112, 15, 113)$.
5. Hence, the required triples are generated.

**Answer:** (24, 7, 25), (40, 9, 41), (60, 11, 61), (84, 13, 85), and (112, 15, 113)

> Common mistake: Confusing the nth odd number with the value of $n$.

### Question 2

*3 marks · Short answer*

Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

**Solution**

1. The method uses the equation $(n - 1)^2 + (2n - 1) = n^2$, where one smaller side is $n - 1$ and the hypotenuse is $n$.
2. Since the difference between the hypotenuse $n$ and the side $n - 1$ is $1$, they cannot share any common factor greater than $1$.
3. Therefore, all triples generated by this method are primitive Baudhāyana triples.
4. Hence, this method does not yield non-primitive triples.

**Answer:** No, this method only yields primitive Baudhāyana triples because the hypotenuse and one smaller side differ by 1.

> Common mistake: Assuming generated triples can be scaled.

### Question 3

*3 marks · Short answer*

Are there primitive triples that cannot be obtained through this method? If yes, give examples.

**Solution**

1. Yes, there are primitive triples where the difference between the hypotenuse and the smaller side is not 1.
2. Consider the primitive triple $(5, 12, 13)$.
3. In $(5, 12, 13)$, the difference between the hypotenuse $13$ and the smaller side $5$ is $8$, not $1$.
4. Therefore, $(5, 12, 13)$ cannot be obtained through this specific method.

**Answer:** Yes, for example, the primitive triple (5, 12, 13) cannot be obtained through this method.

> Common mistake: Failing to check the difference between the sides for given triples.

## Page 52 In-text Questions

### Question 1

*3 marks · Short answer*

“In a lake surrounded by chakra and krauñcha birds, there is a lotus flower peeping out of the water, with the tip of its stem 1 unit above the water. On being swayed by a gentle breeze, the tip touches the water 3 units away from its original position. Quickly tell the depth of the lake.”

**Solution**

1. Let the depth of the lake be $x$ units, so the length of the lotus stem is $x + 1$ units.
2. When the tip touches the water 3 units away from its original position, it forms a right-angled triangle with sides $3$, $x$, and $x + 1$.
3. Using the Baudhayana-Pythagoras theorem, we have $3^2 + x^2 = (x + 1)^2$, which gives $9 + x^2 = x^2 + 2x + 1$, leading to $2x = 8$ and $x = 4$.

**Answer:** The depth of the lake is 4 units.

> Common mistake: Taking the total length of the stem as $x$ instead of $x + 1$.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the diagonal of a square with sidelength 5 cm.

**Solution**

1. Let the sidelength of the square be $a = 5~\text{cm}$.
2. Using the relation for the diagonal of a square, $c = a\sqrt{2}$.
3. Substituting $a = 5$, the diagonal is $5\sqrt{2}~\text{cm}$.

**Answer:** $5\sqrt{2}~\text{cm}$

> Common mistake: Multiplying the sidelength by 2 instead of $\sqrt{2}$.

### Question 2

*3 marks · Short answer*

Find the missing sidelengths in the following right triangles:

**Part (i)**

1. Let the shorter sides be $a = 7$ and $b = 9$, and the hypotenuse be $c$.
2. Using Baudhāyana's Theorem, $c^2 = a^2 + b^2$.
3. $c^2 = 7^2 + 9^2 = 49 + 81 = 130$.
4. Therefore, $c = \sqrt{130}$.

Answer (i): $c = \sqrt{130}$

**Part (ii)**

1. Let the shorter sides be $a = 4$ and $b$, and the hypotenuse be $c = 10$.
2. Using Baudhāyana's Theorem, $a^2 + b^2 = c^2$.
3. $4^2 + b^2 = 10^2 \implies 16 + b^2 = 100$.
4. $b^2 = 100 - 16 = 84$, so $b = \sqrt{84}$.

Answer (ii): $b = \sqrt{84}$

**Part (iii)**

1. Let the shorter sides be $a = 40$ and $b = 41$, and the hypotenuse be $c$.
2. Using Baudhāyana's Theorem, $c^2 = 40^2 + 41^2$.
3. $c^2 = 1600 + 1681 = 3281$, so $c = \sqrt{3281}$.

Answer (iii): $c = \sqrt{3281}$

**Part (iv)**

1. Let the shorter sides be $a = 10$ and $b$, and the hypotenuse be $c = \sqrt{200}$.
2. Using Baudhāyana's Theorem, $10^2 + b^2 = (\sqrt{200})^2$.
3. $100 + b^2 = 200 \implies b^2 = 100$.
4. Therefore, $b = 10$.

Answer (iv): $b = 10$

**Part (v)**

1. Let the shorter sides be $a = 27$ and $b = 45$, and the hypotenuse be $c$.
2. Using Baudhāyana's Theorem, $c^2 = 27^2 + 45^2$.
3. $c^2 = 729 + 2025 = 2754$, so $c = \sqrt{2754}$.

Answer (v): $c = \sqrt{2754}$

**Part (vi)**

1. Let the shorter sides be $a = 10$ and $b$, and the hypotenuse be $c = \sqrt{150}$.
2. Using Baudhāyana's Theorem, $10^2 + b^2 = (\sqrt{150})^2$.
3. $100 + b^2 = 150 \implies b^2 = 50$.
4. Therefore, $b = \sqrt{50}$.

Answer (vi): $b = \sqrt{50}$

**Answer:** See sub-parts for the missing sidelengths.

> Common mistake: Adding or subtracting the given squares incorrectly when finding the missing side.

## Page 53 In-text Questions

### Question 3

*3 marks · Short answer*

Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

**Solution**

1. The diagonals of a rhombus bisect each other at right angles.
2. Let the diagonals be $d_1 = 24$ units and $d_2 = 70$ units, so half-diagonals are $\frac{24}{2} = 12$ units and $\frac{35}{2} = 35$ units.
3. By Baudhāyana's theorem, the side $s$ of the rhombus satisfies $s^2 = 12^2 + 35^2 = 144 + 1225 = 1369$.
4. Thus, $s = \sqrt{1369} = 37$ units.

**Answer:** 37 units

> Common mistake: Using the full lengths of the diagonals instead of half-lengths when applying the theorem.

### Question 4

*3 marks · Short answer*

Is the hypotenuse the longest side of a right triangle? Justify your answer.

**Solution**

1. Yes, the hypotenuse is always the longest side of a right triangle.
2. By Baudhāyana's theorem, if the shorter sides are $a$ and $b$ and the hypotenuse is $c$, then $c^2 = a^2 + b^2$.
3. Since $a$ and $b$ are positive lengths, $c^2$ is greater than both $a^2$ and $b^2$, which means $c$ is greater than both $a$ and $b$.

**Answer:** Yes, because the square of the hypotenuse is the sum of the squares of the other two positive sides.

> Common mistake: Stating yes without referring to the relation of squares of sides.

### Question 5

*1 mark · True or false*

True or False—Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

**Solution**

1. Every Baudhāyana triple is either a primitive triple or obtained by multiplying a primitive triple by a positive integer $k$, which is a scaled version.

**Answer:** True

> Common mistake: Confusing non-primitive triples with triples that cannot be reduced to primitive ones.

### Question 6

*3 marks · Short answer*

Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

**Solution**

1. A rectangle with integer sides $a$ and $b$ has a diagonal $c$ given by $\sqrt{a^2 + b^2}$.
2. Using Baudhāyana triples $(a, b, c)$, the diagonal will also be an integer.
3. Five examples of rectangles are with sides (3, 4), (5, 12), (8, 15), (7, 24), and (12, 35).

**Answer:** Rectangles with sides: (3, 4), (5, 12), (8, 15), (7, 24), and (12, 35)

> Common mistake: Choosing sides that do not form integer diagonals.

### Question 7

*3 marks · Short answer*

Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

**Solution**

1. We need a square whose area equals $7^2 - 5^2$.
2. By rewriting this as $5^2 + c^2 = 7^2$, we can think of $7$ as the hypotenuse and $5$ as one of the shorter sides of a right triangle.
3. The other side $c$ is given by $c^2 = 7^2 - 5^2 = 49 - 25 = 24$, so $c = \sqrt{24}$ units.

**Answer:** A square of sidelength $\sqrt{24}$ units

> Common mistake: Adding the areas instead of subtracting them.

### Question 8

*3 marks · Short answer*

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. unit?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

**Part (i)**

1. For area 2: connect opposite corners of a $1 \times 1$ grid square to form a tilted square of side $\sqrt{2}$, giving area 2.
2. For area 3: it is not possible to create a square of area 3 directly from standard grid dots using simple integer steps.
3. For area 4: connect dots spanning $2$ units horizontally and vertically to form a square of area $2^2 = 4$.
4. For area 5: form a square using sides that are diagonals of $1 \times 2$ rectangles ($1^2 + 2^2 = 5$).

Answer (i): (a) Possible (b) Not possible (c) Possible (d) Possible

**Part (ii)**

1. Any square on a grid can be formed by a right-angled triangle with integer legs $x$ and $y$ such that its area is $x^2 + y^2$.
2. Thus, the possible integer-valued areas are sums of two square numbers.

Answer (ii): Numbers that can be expressed as the sum of two integer squares

**Answer:** See individual parts.

> Common mistake: Assuming all integer areas can be formed on a grid.

## Page 54 In-text Questions

### Question 9

*3 marks · Short answer*

Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

**Solution**

1. Let the equilateral triangle have side $s = 6$ units, and draw an altitude from one vertex to the opposite side.
2. The altitude bisects the opposite side into two equal halves of length $\frac{6}{2} = 3$ units, forming a right-angled triangle with hypotenuse $6$ units and one side $3$ units.
3. Using the Baudhayana-Pythagoras theorem, if $h$ is the height, then $3^2 + h^2 = 6^2$, which gives $9 + h^2 = 36$, so $h^2 = 27$ and $h = \sqrt{27} = 3\sqrt{3}$ units.
4. The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}$ sq. units.

**Answer:** $9\sqrt{3}$ sq. units

> Common mistake: Taking the full side length instead of half the base when applying the theorem for the height of the triangle.

## Frequently asked questions

### How many total questions are covered in the NCERT Solutions for Class 8 Maths Chapter 9: The Baudhayana-Pythagoras Theorem?

The solutions cover all in-text questions and Figure It Out sections from the new NCERT book for the 2026-27 session. You can find comprehensive step-by-step explanations for every single question right here on this page.

### Which major topics and concepts are included in this chapter?

This chapter covers concepts like doubling and halving a square, decimal and lower-upper bounds of $\sqrt{2}$, Baudhāyana-Pythagoras theorem applications, and generating Baudhāyana triples. All these topics are thoroughly explained in SwaVid's free PDF and solutions on this page.

### Which type of questions are considered the most challenging in this chapter?

Questions involving the generation of primitive Baudhāyana triples, odd square methods, and finding bounds for $\sqrt{2}$ are often found tricky by students. To approach them, you should carefully follow the geometric proofs and grid area visualizations provided in our step-by-step solutions.

### How should I write my answers to score full marks in Class 8 exams?

To score full marks, your answers must clearly state the geometric properties used, such as triangle congruence, diagonal symmetry, or the application of the Baudhāyana-Pythagoras theorem. SwaVid's solutions provide structured formats that help you write precise steps and reasoning for every problem.

### Is a free PDF of these solutions available for download?

Yes, the complete and free PDF containing step-by-step solutions for Class 8 Maths Chapter 9 is available on this page only. You can easily access or download it to study offline for the 2026-27 academic session.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student